Giant Numbers and Modulo Arithmetic

A: At first blush this is an impossible task. The above number is impossible to compute. However, using some simple rules from modulo arithmetic, it is possible to answer this.

Let us first try to solve this the long way and run some experiments with 7.
$$
7^{2} = 49 \text{ last digit is 9}\\
7^{3} \equiv 49 \times 7 \equiv \text{ last digit is 3 as } 9\times7 \equiv \text{ last digit is 3}\\
$$
Continuing in the same fashion
$$
7^{4} \equiv 3 \times 7 \equiv \text{ last digit is 1 as } 3 \times 7 = 21
$$
Now \(7 ^ {5}\) would have its last digit as 7, because 21's last digit is 1 and when multiplied by 7 yields 7. But this is the same as \(7^{1}\). This resets the entire process. We can conclude that as we multiply into higher powers of 7, the last digits cycles (of length 4) as \({7,9,3,1,...}\). This implies, \(7^{2000}\) would have the units place as 1, as 2000 divides 4 with no remainders.

The problem can be solved in a terser way if we use rules from modulo arithmetic. A refresher on modulo arithmetic. When we write \( a \equiv b (\textrm{mod}\ c)\) we mean that \(a-b\) is divisible by \(c\) without leaving any remainder.

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